A naive question about the scheme theory: regard $\mathbb C^n$ as a scheme Note that $\mathbb C$ can be regarded as the set of closed points of $\mathrm{Spec} ~\mathbb C[T]$. And, $\mathbb C^n$ should be regarded as that of $\mathrm{Spec} ~\mathbb C[T_1,\cdots,T_n]$
What if we replace $\mathbb C$ here by a general ring $R$? What can we say about a ring $R$ (resp. $R^{\oplus n}$)and the set of closed points in $\mathrm{Spec}R[T]$(resp. $\mathrm{Spec}R[T_1,\cdots,T_n]$)?
Specifically, under what conditions on a given ring $R$, can we find another ring $S$ so that $R$ can be identified with (the set of closed points of) the affine scheme $\mathrm{Spec}(S)$? For example, is it necessary that $R$ is a field?
 A: In general, spectrum of polynomial rings $R[T_1, \cdots, T_n]$ are much more complicated compared to when $R$ is an algebraically closed field like $\mathbb{C}$. Let me demonstrate by a few sequentially escalating examples. Recall that a closed point of an affine scheme $\mathrm{Spec}\: A$ correspond to maximal ideals of $A$. 


*

*Example, $\mathbb{R}[T]$: What are the closed points of $\mathrm{Spec}\: \mathbb{R}[T]$. This is the closest thing to $\mathbb{C}[T]$, right? What are the maximal ideals of $\mathbb{R}[T]$. There are two types of irreducible polynomials here
$$
p_a(T)=T-a, \qquad q_{\alpha}(T)=(T-\alpha)(T-\overline{\alpha})
$$
where in the above $a\in \mathbb{R}$ and $\alpha\in \mathbb{C}$ is such that $\mathrm{Im}\: \alpha \neq 0$. As a result, the set of closed points of $\mathbb{R}[T]$ (thought as a subset of complex plane) is
$$\boxed{\mathrm{CP}(\mathrm{Spec}\: \mathbb{R}[T])\simeq \mathbb{R}\cup \mathbb{H}}$$ where $\mathbb{H}$ is the upper half plane ($\mathrm{Im}\: z>0)$. As you can see, the closed points are much more than just $\mathbb{R}$.

*Example, $\mathbb{F}_p[T]$: Let $\mathbb{F}_p$ be a finite field with $p$ a prime number (for simplicity we are sticking to $p$ and not a power of it). What are the closed points of $\mathbb{F}_p[T]$? Note that we are still sticking with the base ring a field, but as you'll see the set of closed points here is even more complicated. Let $P(T)\in \mathbb{F}_p[T]$ be a monic irreducible polynomial of degree $n$. This polynomial then defines a field extension of $\mathbb{F}_p$, namely $\mathbb{F}_{p^n}$, such that $\alpha\in \mathbb{F}_{p^n}$ is a root of $P(T)$ and moreover
$$
P(T) = (T-\alpha)(T-\alpha^p)(T-\alpha^{p^2})\cdots (T-\alpha^{p^{n-1}})
$$
with $\alpha^{p^s}\in \mathbb{F}_{p^n}$ for all $0\leq s\leq n-1$. The algebraic closure of $\mathbb{F}_p$ is the union
$
\overline{\mathbb{F}_p} = \bigcup_{n=1}^\infty \mathbb{F}_{p^n}
$.
Given $\alpha, \beta\in \mathbb{F}_{p^n}$, define the equivalence relation $\alpha\sim \beta$ iff $\beta=\alpha^{p^m}$ for some $m$. Then the set of closed points of $\mathbb{F}_p[T]$ becomes the quotient space $$\boxed{\mathrm{CP}(\mathrm{Spec}\: \mathbb{F}_p[T])\simeq \overline{\mathbb{F}_p}/\sim}$$ Well, it least $\mathbb{F}_p\subset \overline{\mathbb{F}_p}/\sim$. We don't even have that in the rest of the examples!

*Example, $\mathbb{Z}[T]$: Enough with fields though, let's take a look at $\mathbb{Z}[T]$; what are its closed points? The maximal ideals of this ring are of the form $(p, f(T))$ with $p$ a prime number and $f(T)$ an irreducible polynomial modolu $p$. i.e.
$$
\boxed{\mathrm{CP}(\mathrm{Spec}\: \mathbb{Z}[T])\simeq \{(p,f(T)\mid p\text{ prime}, f(T) \text{ irreducible in }\mathbb{F}_p[T]\}}
$$
As you can see, there is no clear connection between this set and $\mathbb{Z}$.

*Example, $\mathbb{C}[X][T]$: What if we take $R=\mathbb{C}[X]$ and look at $\mathbb{C}[X][T]$? Well, $\mathbb{C}[X][T]=\mathbb{C}[T,X]$. The closed points of the latter are in correspondence with 
$$\boxed{\mathrm{CP}(\mathrm{Spec}\: \mathbb{C}[X][T])\simeq \mathbb{C}^2}$$
Note that $\mathbb{C}^2$ is 2-dimensional as a $\mathbb{C}$-vector space. However, the base ring $\mathbb{C}[X]$, is infinite dimensional as a $\mathbb{C}$ vector space.
I hope by now you can see how terribly complicated and different things can get depending on what the base ring $R$ is for a polynomial ring $R[T_1, \cdots, T_n]$, when you are interested in closed points. Also, note that in my all of my examples the base ring $R$ is a Noetherian integral domain. Also, all of my examples are of the form $R[T]$ (just one variable). In a way, all of my examples are "nice, well-behaving and simple". Things can get much much worse than this!
